1. If \(\sin \theta = \cfrac{p^2 - q^2}{p^2 + q^2}\), then show that \(\cot \theta = \cfrac{2pq}{p^2 - q^2}\) where \(p > q\) and \(0^\circ < \theta < 90^\circ\).

2. If \(\tan\theta = \cos 30^\circ + \sin 60^\circ\), then \(\sin(90^\circ - \theta)\) will be –

(a) \(\cfrac{1}{3}\) (b) \(\cfrac{1}{4}\) (c) \(\cfrac{1}{2}\) (d) \(1\)

3. Given that \(\sin(A + B) = 1\) and \(\cos(A - B) = 1\), where \(0^\circ \leq A + B \leq 90^\circ\) and \(A > B\), determine the values of angles \(A\) and \(B\).

4. If \(0^\circ < \theta < 90^\circ\), then show that \(\sin\theta + \cos\theta > 1\)

5. If \(x^2 = \sin^2 30^\circ + 4\cot^2 45^\circ - \sec^2 60^\circ\), then the value of \(x\) is —

(a) \(\pm 1\) (b) \(\pm \cfrac{1}{2}\) (c) \(\pm \cfrac{1}{\sqrt2}\) (d) \(\pm \cfrac{1}{\sqrt3}\)

6. Evaluate the expression: \[ \frac{4}{3} \cot^2 30^\circ + 3 \sin^2 60^\circ - 2 \csc^2 60^\circ - \frac{3}{4} \tan^2 30^\circ \]

7. Evaluate the expression: \[ \cot^2 30^\circ - 2\cos^2 60^\circ - 4\sin^2 30^\circ - \frac{3}{4}\sec^2 45^\circ + \tan 45^\circ \]

8. If \(x = \sin^2 30^\circ + 4 \cot^2 45^\circ - \sec^2 60^\circ\), find the value of \(x\).

9. Solve: \[ x^2 = \sin^2 30^\circ + 4\cot^2 45^\circ - \sec^2 60^\circ \]

10. If \(a(\tan\theta + \cot\theta) = 1\) and \(\sin\theta + \cos\theta = b\), then prove that \(2a = b^2 - 1\), where \(0^\circ < \theta < 90^\circ\).

11. Find the value of \(x\) from the equation: \(x \sec^2 45^\circ \cdot \csc^2 45^\circ + 2(\sin 60^\circ + \sin 30^\circ) = \tan 60^\circ\)

12. \(\sin(2x-20°)=\cos(2y+20°)\)
বা, \(\sin(2x-20°)=\sin[90^o-(2y+20°)]\)
বা, \(2x-20°= 90^o-(2y+20°)\)
বা, \(2x-20°= 90^o-2y-20°\)
বা, \(2x+2y= 90^o-20°+20°\)
বা, \(2(x+y)= 90^o\)
বা, \((x+y)= 45^o\)

\(\therefore \tan(x+y)=\tan 45^o=1\) (Answer)
- translate in english

13. Evaluate the expression: \[ \frac{1 - \sin^2 30^\circ}{1 + \sin^2 45^\circ} \times \frac{\cos^2 60^\circ + \cos^2 30^\circ}{\csc^2 90^\circ - \cot^2 90^\circ} \div (\sin 60^\circ \tan 30^\circ) \]

14. Evaluate the expression: \[ \frac{(\sin 0^\circ + \sin 60^\circ)(\cos 60^\circ + \cot 45^\circ)}{(\cot 60^\circ + \tan 30^\circ)(\csc 30^\circ - \csc 90^\circ)} \]

15. If \(x^2 = \sin^2 30^\circ + 4\cot^2 45^\circ - \sec^2 60^\circ\), then what is the value of \(x\)?

16. If \(\sin A + \sin B = 2\), where \(0^\circ \leq A \leq 90^\circ\) and \(0^\circ \leq B \leq 90^\circ\), then find the value of \(\cos A + \cos B\).

17. Determine the values of \(\theta\) for which \(\sin^2\theta - 3\sin\theta + 2 = 0\) holds true, given that \(0^\circ < \theta < 90^\circ\).

18. If \(x^2 = \sin^2 30^\circ + 4\cot^2 45^\circ - \sec^2 60^\circ\), find the value of \(x\). Let me know if you'd like a step-by-step solution next. I'm ready when you are.

19. If \(x^2 = \sin^2 30^\circ + 4\cot^2 45^\circ - \sec^2 60^\circ\), find the value of \(x\).

20. Find the value of: \( \cfrac{4}{3} \cot^2 30^\circ + 3 \sin^2 60^\circ - 2 \csc^2 60^\circ - \cfrac{3}{4} \tan^2 30^\circ \)

21. Find the value of: \( \cfrac{\cfrac{1}{3} \cos 30^\circ}{\cfrac{1}{2} \sin 45^\circ} + \cfrac{\tan 60^\circ}{\cos 30^\circ} \)

22. Find the value of: \( \cfrac{\tan 60^\circ - \tan 30^\circ}{1 + \tan 60^\circ \tan 30^\circ} + \cos 60^\circ \cos 30^\circ + \sin 60^\circ \sin 30^\circ \)

23. Find the value of: \( \cfrac{1 - \sin^2 30^\circ}{1 + \sin^2 45^\circ} \times \cfrac{\cos^2 60^\circ + \cos^2 30^\circ}{\csc^2 90^\circ - \cot^2 90^\circ} \div (\sin 60^\circ \cdot \tan 30^\circ) \)

24. If \(x^2 = \sin^2 30^\circ + 4\cot^2 45^\circ - \sec^2 60^\circ\), then find the value of \(x\).

25. If \(\sin(A + B) = 1\) and \(\cos(A - B) = 1\), then find the value of \(\cot 2A\), given that \(0^\circ \leq (A + B) \leq 90^\circ\) and \(A \geq B\).

26. If \(\sec \theta = \csc \phi\), where \(0^\circ < \theta < 90^\circ\) and \(0^\circ < \phi < 90^\circ\), then the value of \(\sin(\theta + \phi)\) is 1.

27. ```html \(\cfrac{\sinθ}{x}=\cfrac{\cosθ}{y}\)
i.e., \(\cfrac{\sinθ}{\cosθ}=\cfrac{x}{y}\)
i.e., \(\tanθ=\cfrac{x}{y}\)
i.e., \(\tan^2θ=\cfrac{x^2}{y^2}\)
i.e., \(1+ \tan^2θ=1+\cfrac{x^2}{y^2}\)
i.e., \(\sec^2θ=\cfrac{y^2+x^2}{y^2}\)
i.e., \(\secθ=\cfrac{\sqrt{y^2+x^2}}{y}\)
i.e., \(\cosθ=\cfrac{y}{\sqrt{x^2+y^2}} \)

Now, in the equation \(\cfrac{\sinθ}{x}=\cfrac{\cosθ}{y}\),
substituting \(\cosθ=\cfrac{y}{\sqrt{x^2+y^2}}\), we get
\(\cfrac{\sinθ}{x}=\cfrac{\cfrac{y}{\sqrt{x^2+y^2}}}{y}\)
i.e., \(\cfrac{\sinθ}{x}=\cfrac{1}{\sqrt{x^2+y^2}}\)
i.e., \(\sinθ=\cfrac{x}{\sqrt{x^2+y^2}}\)

∴ \(\sinθ−\cosθ=\cfrac{x}{\sqrt{x^2+y^2}}−\cfrac{y}{\sqrt{x^2+y^2}}\)
\(=\cfrac{x−y}{\sqrt{x^2+y^2}}\) (Proved) ```

28. \[ \tan 60^\circ = \sqrt{3},\quad \tan 30^\circ = \frac{1}{\sqrt{3}} \\ \cos 60^\circ = \frac{1}{2},\quad \cos 30^\circ = \frac{\sqrt{3}}{2} \\ \sin 60^\circ = \frac{\sqrt{3}}{2},\quad \sin 30^\circ = \frac{1}{2} \] \[ \frac{\tan 60^\circ - \tan 30^\circ}{1 + \tan 60^\circ \cdot \tan 30^\circ} = \frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + \sqrt{3} \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{3 - 1}{\sqrt{3}}}{2} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}} \] \[ \cos 60^\circ \cos 30^\circ + \sin 60^\circ \sin 30^\circ = \frac{1}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \cdot \frac{1}{2} = \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2} \] \[ \frac{1}{\sqrt{3}} + \frac{\sqrt{3}}{2} = \frac{2 + 3}{2\sqrt{3}} = \frac{5}{2\sqrt{3}} \]

29. If \(\alpha + \beta = 90^\circ\) and \(\alpha : \beta = 2 : 1\), then find the value of \(\sin \alpha : \sin \beta\).

(a) \(3:1\) (b) \(1:3\) (c) \(\sqrt3:1\) (d) \(1:\sqrt 3\)

30. Evaluate the value of: \(\cfrac{\tan60^\circ - \tan30^\circ}{1 + \tan60^\circ \cdot \tan30^\circ} + \cos60^\circ \cdot \cos30^\circ + \sin60^\circ \cdot \sin30^\circ\)"